University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 2 - Section 2.3 - The Precise Definition of a Limit - Exercises - Page 76: 23


1) The interval is $(-2.1213,-1.8708)$ 2) $\delta=0.1213$

Work Step by Step

Find a $\delta\gt0$ such that for all $x$ $$0\lt |x-(-2)|\lt\delta\Rightarrow|(x^2)-4|\lt0.5$$ 1) Find the interval around $-2$ on which $|(x^2)-4|\lt0.5$ holds. Solve the inequality: $$|(x^2)-4|\lt0.5$$ $$-0.5\lt x^2-4\lt0.5$$ $$3.5\lt x^2\lt4.5$$ Now we need to take the square root. However, there are 2 cases, whether $x\ge0$ or $x\lt0$. - For $x\ge0$: $1.8708\lt x\lt2.1213$ - For $x\lt0$: $-2.1213\lt x\lt-1.8708$ Yet, here we are looking for the interval around $-2$, which means $x\lt0$. Therefore, we choose only the open interval around $-2$, which is $(-2.1213,-1.8708)$. 2) Give a value for $\delta$ - From $-2.1213$ to $-2$: $|-2.1213|-|-2|=0.1213$ - From $-1.8708$ to $-2$: $|-2|-|-1.8708|=0.1292$ The nearer endpoint to $-2$ is $-2.1213$, and the distance between them is $0.1213$. So if we take $\delta=0.1213$ or any smaller positive number, then $0\lt|x-(-2)|\lt0.1213$, meaning all $x$ would be placed in the interval $(-2.1213,-1.8708)$ so that $|(x^2)-4|\lt0.5$. In other words, $$0\lt |x-(-2)|\lt0.1213\Rightarrow|(x^2)-4|\lt0.5$$
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